Aloha :)
$$f(x)=\frac{2}{x}\quad;\quad x_0=2$$Wir bilden zunächst den Differenzenquotienten$$\frac{f(x_0+h)-f(x_0)}{h}=\frac{\frac{2}{x_0+h}-\frac{2}{x_0}}{h}=\frac{\frac{2x_0}{x_0(x_0+h)}-\frac{2(x_0+h)}{x_0(x_0+h)}}{h}=\frac{\frac{2x_0-2(x_0+h)}{x_0(x_0+h)}}{h}$$$$\quad=\frac{2x_0-2(x_0+h)}{h\,x_0(x_0+h)}=\frac{2x_0-2x_0-2h}{h\,x_0(x_0+h)}=\frac{-2h}{h\,x_0(x_0+h)}=\frac{-2}{x_0(x_0+h)}$$und lassen nun \(h\to0\) laufen:
$$f'(x_0)=\lim\limits_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}=\lim\limits_{h\to0}\frac{-2}{x_0(x_0+h)}=-\frac{2}{x_0^2}$$$$f'(2)=-\frac{2}{2^2}=-\frac{1}{2}$$